cancel out the common terms and factor out the common terms and extract the coefficients

Question

the following code fail to multiply each equation by dt, since dt does not seem to do any jobs. it remains as a common factor, I would like it to cancel out the 1/dt term and also multiply the rest of them with dt.

eqns = {x1'[t] == (\[Mu] - (x1^2 + x2^2)) x1 - x3x2 + 
    Ksin + \[Sigma]dw/dt, 
  x2'[t] == (\[Mu] - (x1^2 + x2^2)) x2 + (x3x1), 
  x3'[t] == -(Ksin + \[Sigma]dw/dt) x2}

sort1 = eqns /. {x1'[t] -> dx1/dt, x2'[t] -> dx2/dt, 
    x3'[t] -> dx3/dt} // MatrixForm

sort2 = sort1*dt

what I would like to see is

dx1 == dt Ksin - dt x1^3 - dt x1 x2^2 - dt x3x2 + 
  dt x1 \[Mu] + \[Sigma]dw
dx2 == -dt x1^2 x2 - dt x2^3 + x3x1dt + dt x2 \[Mu]
dx3 == -x2 (dt Ksin + \[Sigma]dw)

besides,if it works as what I expect, how can you extract the terms/coefficients that includes the dt and dw seperately? it should look like the following expressions

dx1=dt(Ksin-x1^3-x1x2^2-x3x2+x1\[Mu])+(\[Sigma])dw)
dx2 == dt(-x1^2 x2 - x2^3 + x3x1 + x2 \[Mu])
dx3 == -dt(Ksin) + (\[Sigma])dw

if possible, can we make them as a matrix form?

since I would like to use the following terms elsewhere to do some other analysis

Ksin-x1^3-x1x2^2-x3x2+x1\[Mu]
-x1^2 x2 - x2^3 + x3x1 + x2 \[Mu]

Sincerely, Li

Answer 1

EDIT: Better yet, use MultiplySides (which I learned about from this answer).

eqns = {x1'[t] == (\[Mu] - (x1^2 + x2^2)) x1 - x3 x2 + Ksin + \[Sigma]dw/dt,
        x2'[t] == (\[Mu] - (x1^2 + x2^2)) x2 + x3 x1,
        x3'[t] == -(Ksin + \[Sigma] dw/dt) x2};
sort = First@Solve[eqns, {x1'[t], x2'[t], x3'[t]}] /. Rule -> Equal;
sort2 = MultiplySides[sort, dt, Assumptions -> dt != 0];
sort3 = sort2 /. {x1'[t] -> dx1/dt, x2'[t] -> dx2/dt, x3'[t] -> dx3/dt} // TableForm

which produces

{
 {dx1 == dt Ksin - dt x1^3 - dt x1 x2^2 - dt x2 x3 + dt x1 \[Mu] + \[Sigma]dw},
 {dx2 == dt (-x1^2 x2 - x2^3 + x1 x3 + x2 \[Mu])},
 {dx3 == -x2 (dt Ksin + dw \[Sigma])}
}

To extract coefficients do the following

beloweqns1 = {dx1/dt == (a + b + c) dt + (e + f) dw, 
              x2/dt == (a1 + a2 + a3) dt + (e1 + e2) dw};
Coefficient[beloweqns1[[1, 2]], dt]
Coefficient[beloweqns1[[1, 2]], dw]
Coefficient[beloweqns1[[2, 2]], dt]
Coefficient[beloweqns1[[2, 2]], dw]

To understand the indexing on beloweqns above, use TreeForm[beloweqns].

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I think you want to use

Distribute[sort1*dt, Equal]

because otherwise Mathematica treats the equation as a symbolic object.